Problem: $f(x) = \begin{cases} \dfrac{ 1 }{ \sqrt{ x - 2 } } & \text{if } x \geq 2 \\ \dfrac{ 1 }{ \sqrt{ 2 - x } } & \text{if } x < 2 \end{cases}$ What is the domain of the real-valued function $f(x)$ ?
Solution: $f(x)$ is a piecewise function, so we need to examine where each piece is undefined. The first piecewise definition of $f(x)$ $\frac{ 1 }{ \sqrt{ x - 2 } }$ , is undefined where the denominator is zero and where the radicand (the expression under the radical) is less than zero. The denominator, $\sqrt{ x - 2 }$ , is zero when $x - 2 = 0$ , so we know that $x \neq 2$ The radicand, $x - 2$ , is less than zero when $x < 2$ , so we know that $x \geq 2$ So the first piecewise definition of $f(x)$ is defined when $x \neq 2$ and $x \geq 2$ . Combining these two restrictions, the first piecewise definition is defined when $x > 2$ . The first piecewise defintion applies when $x \geq 2$ , so this restriction is relevant. The second piecewise definition of $f(x)$ $\frac{ 1 }{ \sqrt{ 2 - x } }$ , applies when $x < 2$ and is undefined where the denominator is zero and where the radicand is less than zero. The denominator, $\sqrt{ 2 - x }$ , is zero when $2 - x = 0$ , so we know that $x \neq 2$ The radicand, $2 - x$ , is less than zero when $x > 2$ , so we know that $x \leq 2$ So the second piecewise definition of $f(x)$ is defined when $x \neq 2$ and $x \leq 2$ . Combining these two restrictions, the second piecewise definition is defined when $x < 2$ . However, the second piecewise definition of $f(x)$ only applies when $x < 2$ , so restriction isn't actually relevant to the domain of $f(x)$ So the first piecewise definition is defined when $x > 2$ and applies when $x \geq 2$ ; the second piecewise definition is defined when $x < 2$ and applies when $x < 2$ . Putting the restrictions of these two together, the only place where a definition applies and the value is undefined is at $x = 2$ . So the only restriction on the domain of $f(x)$ is $x \neq 2$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \neq2\, \}$.